Implicit Differentiation

The techniques we learned allows us to differentiate functions like `y=root(3)(x^2+sin(x))` or `y=tan(x)/(e^(sin(x)))` or, in general, `y=f(x)`. But what to do if we can't express `y` in terms of `x`, or expression will be very complex?

Consider, `x^2+y^2=16`. This is equation of circle and we can solve for `y`: `y=+-sqrt(16-x^2)`. But here we obtained two functions and we have more work than earlier.

For example, `x^3+y^3=2xy` can be solved for `y`, but expression will be very complex and finally `x^2+y^4=sin(x+y)` can't be solved explicitly for `x`.

What to do in such cases?

Fortunately, we don't need to express `y` in terms of `x`. We can use implicit differentiation. This consists of differentiating both sides of the equation with respect to `x` and then solving the resulting equation for `y'`. We assume that the given equation determines `y` implicitly as a differentiable function of `x` so that the method of implicit differentiation can be applied.

Example 1. Find equation of tangent line to `x^2+y^2=25` at point (3,4).

To find slope of tangent line we need derivative of function. Here function is given implicitly, so we use implicit differentiation.

Differentiate both sides of equation with respect to `x` and remember that `y` is a function of `x`.

`d/(dx)(x^2+y^2)=d/(dx)(25)`

`d/(dx)(x^2)+d/(dx)(y^2)=0`

`2x+2y(dy)/(dx)=0`

Notice how we used chain rule while differentiating `y^2` : `d/(dx)(y^2)=d/(dy)(y^2)*(dy)/(dx)=2y(dy)/(dx)`.

Solving for `(dy)/(dx)` gives `(dy)/(dx)=-x/y` .

Therefore, at point (3,4) slope of tangent line is `-3/4` and equation of tangent line is `y-4=-3/4(x-3)` or `y=-3/4x+25/4`.

As can be seen implicit differentiation can make differentiation easier and can save load of time.

Example 2. Where tangent line is horizontal or vertical to `x^3+y^3=3xy` (folium of Descartes).

Differentiating both sides of equation with respect to `x` we obtain

`d/(dx)(x^3+y^3)=d/(dx)(3xy)` or

`d/(dx)(x^3)+d/(dx)(y^3)=3d/(dx)(xy)`

To differentiate `xy` we need to use product rule: `d/(dx)(xy)=d/(dx)(x)y+xd/(dx)(y)=y+x(dy)/(dx)`

To differentiate `y^3` we need to use chain rule: `d/(dx)y^3=d/(dy)(y^3)*(dy)/(dx)=3y^2(dy)/(dx)`.

So, equation can be rewritten as

`3x^2+3y^2(dy)/(dx)=3(y+x(dy)/(dx))`.

Solving for `(dy)/(dx)` gives `(dy)/(dx)=(y-x^2)/(y^2-x)`.

Tangent line is horizontal when `(dy)/(dx)=0` or `y-x^2=0` which gives `y=x^2`.

If we now substitute `y` in initial equation with obtained expression, we will have that `x^3+( x^2)^3=3x*x^2` or `x^6-2x^3=0`.

Thus, tangent line is horizontal either when `x=0` or `x=2^(1/3)`.

So, tangent line is horizontal at points (0,0) and `(2^(1/3),(2^(1/3))^2)=(2^(1/3),2^(2/3))~~(1.26,1.59)`.

The tangent line is vertical when the denominator in the expression for `(dy)/(dx)` is 0 or `y^2=x.` Another method is to observe that the equation of the curve is unchanged when `x` and `y` are interchanged, so the curve is symmetric about the line `y=x`. This means that the horizontal tangents at (0, 0) and `(2^(1/3),2^(2/3))` correspond to vertical tangents at (0, 0) and `(2^(2/3),2^(1/3))`.